LC natural response derivation 2

In the last video, we set up this differential equation that described an LC circuit, and now we're going to go about solving this second-order circuit. The technique that works here is the same that worked with first-order ordinary differential equations. We're looking for a function of i that makes this whole equation true, and we're going to do that by guessing at an i, putting it back into the equations, and seeing if it works. If it works, we win; if it doesn't work, we have to think of something else, and we keep doing that until we solve it.

So knowing what I know about derivatives and looking at this equation, can I guess at a plausible solution? What we have here are two terms that have to add up to zero for all time. That means that whatever function I pick for i—and there are some scaling factors here—let me write this equation like this. So what I have here is a scaling factor times i that has to somehow fully subtract from the second derivative of i. These two terms have to somehow look alike for all time.

There's one function I know where its derivatives sort of look like what we started with, and that's the exponential function. So I'm actually going to make a guess. I'm going to guess that i of t is something of the form some constant times the exponential of time with some other scale factor. Now, k is an adjustable parameter, and that's an amplitude, telling us how big the signal is. What's s? s is up in the exponent along with t, and we know that by the time we take an exponent of anything, that whatever's up here has to have no units. So that means that s times t has no units, and that means that s has units of one over time.

So s is a frequency of some sort, and in particular, it's going to be a radian frequency. It's going to be in radians per second, so s is going to be called the natural frequency. Let's keep working on this. We're going to try to plug i back into our equation and see if it works. We can plug i straight into here, and we need the second derivative of i.

So let's first take the first derivative: d/dt of i equals d/dt of k e^(st). We can take that derivative, and that equals k times s times e^(st), so it's also an exponential. Now we need the second derivative, so we want to take the derivative of this guy. The second derivative of i with respect to time equals the derivative of the first derivative of k e^(st), and that equals another s that comes down, so it's s² k e^(st).

Good. Now we have our second derivative; we can plug that in here. So the equation becomes s² k e^(st) + (1/LC) k e^(st) = 0. Let's do a little factoring. There's a common term here: k e^(st). Let me factor that out: k e^(st) times what's left: s² + (1/LC) = 0.

Okay, we have how many adjustable parameters here? We have k and s; those are the two. L and C are constants that are values of our circuit. So we need to find some values of k and s that make this equation zero. I can make k equal to zero, and that would mean that 0 = 0, so our amplitude would be 0. If we put nothing in the circuit, we get nothing out. That's totally boring, so that's not so interesting of a solution.

Now, does e^(st) ever become zero? e to something? It never does. If I let t go to plus infinity and s is negative, then e^(st) would become zero, but plus infinity time from now is pretty far in the future, and I don't want to wait that long. So the interesting solution becomes: can we make s² + (1/LC) equal to zero? This equation is referred to as the characteristic equation.

Let's see what happens when we try to solve this. The first step of this is, I'm going to get s² = -1/LC or s = sqrt(-1/LC). Uh oh, look what happens here; we're taking the square root of a negative number. So what's going to happen here? We're going to get an imaginary number for our answer. I can write this as sqrt(-1) times sqrt(1/LC), and that gives me two answers. I'm going to call the first answer s₁, and that's going to be equal to j, which is the imaginary number (that's the square root of -1 for electrical engineers), times sqrt(1/LC). The other s₂ is the negative of that: -j.

So these are two possible solutions to our differential equation. Now I'm going to give a nickname to this expression here because I don't want to write it so much; I'm going to call this ω₀. This is a lowercase omega, a Greek letter. This is the capital omega that we use for ohms, but little omega is often used as this variable here. So I can say that s = s₁ = +jω₀ and s₂ = -jω₀.

We're going to use this notation for a little bit, but just remember I made that simple substitution. We have two different roots that can cause our differential equation to go to zero. So when we combine these into a solution for i, we're going to say i = k₁ e^(s₁ t) + k₂ e^(s₂ t). I could also write this as, let's fill in some of the numbers here: k₁ e^(+jω₀ t) + k₂ e^(-jω₀ t). So that's my proposed solution.

What we need to do now, we found s; we found two values of s. Now we have these two constants. We got to work those out. So in the next video, we'll use the initial conditions to figure out what k₁ and k₂ are.